Electric Field and Electric Potential of a Point Charge

The electric field of a point charge is E = kq/r^2, a vector that points away from positive charge and toward negative charge. The electric potential is V = kq/r, a scalar with a sign but no direction. Field falls off as 1/r^2, potential as 1/r; both use k = 9.0 x 10^9 N m^2/C^2.

AP Physics: Unit 10 (topics 10.3 Electric Fields, 10.4 Electric Potential Energy, 10.5 Electric Potential). Covers Topics 10.3 through 10.5 of AP Physics 2 Unit 10, Electric Force, Field, and Potential. In AP Physics C: Electricity and Magnetism, the same ideas span Unit 8 (Electric Charges, Fields, and Gauss's Law) and Unit 9 (Electric Potential).

How to find the electric field of a point charge

To find the electric field of a point charge, use $E = \frac{kq}{r^2}$ , where $k = 9.0 \times 10^9\ \mathrm{N \cdot m^2/C^2}$ is the Coulomb constant, $q$ is the charge creating the field, and $r$ is the distance from that charge to the point where you want the field.

The result is a vector. Plug in the magnitude of $q$ to get the magnitude of $E$ , then assign direction by inspection: the field points radially away from a positive charge and radially toward a negative charge.

The field tells you the force a test charge would feel per coulomb, $E = F/q$ . Multiply the field by any charge you place at that point and you recover the force from Coulomb's law, which is why this formula looks like Coulomb's law with one charge stripped out. Units are newtons per coulomb (N/C), which is the same thing as volts per meter (V/m).

The electric potential formula

The electric potential formula for a point charge is $V = \frac{kq}{r}$ , with the same $k$ and the same meaning for $r$ , except $r$ appears to the first power, not squared. Potential is a scalar: it has a sign but no direction, so there is never an angle or a component to worry about.

Unlike the field calculation, you keep the sign of the charge. A positive charge creates positive potential everywhere around it; a negative charge creates negative potential. The formula takes $V = 0$ at $r = \infty$ , the standard reference point for a point charge.

Units are volts, and one volt is one joule per coulomb. That definition is the practical meaning of potential: it tells you how much potential energy each coulomb would have at that location, $V = U_E/q$ , before you have decided what charge, if any, actually sits there.

Field vs potential: the distinction that costs points

Students cross these up constantly because the formulas differ by a single power of $r$ . Keep the comparison straight:

	Field $E$	Potential $V$
Formula	$E = kq/r^2$	$V = kq/r$
Quantity	Vector	Scalar
Units	N/C (equals V/m)	V (equals J/C)
Distance dependence	Falls as $1/r^2$	Falls as $1/r$
Sign of $q$	Sets the direction	Sets the sign of $V$

A quick self-test: if you double your distance from a point charge, the field drops to one quarter of its old value, but the potential only drops to one half. If your answer to a find-the-field question has no direction attached, or your find-the-potential answer somehow has components, you have mixed the two up.

Potential energy and the charge that moves

Potential becomes useful the moment a charge moves. The change in electric potential energy is $\Delta U_E = q\Delta V$ , where $q$ is the charge that moves (sign included) and $\Delta V = V_f - V_i$ is the potential difference it moves through.

Signs do the physics here. A positive charge that moves toward lower potential has negative $\Delta U_E$ : it loses potential energy and, if the electric force is the only force acting, gains kinetic energy. A negative charge does the opposite, speeding up as it moves toward higher potential. Once you have $\Delta U_E$ , the problem becomes ordinary conservation of energy, with $\Delta K = -\Delta U_E$ when no other forces act.

This is why potential earns its own topic. You can answer how-fast-is-it-moving questions without ever computing a force or an acceleration along the way.

Multiple charges: superposition works differently for each

With more than one point charge, both quantities obey superposition, but the bookkeeping differs.

Fields add as vectors. Find each charge's field magnitude with $E = kq/r^2$ , draw each field's direction at the point of interest, then add components.
Potentials add as signed numbers. Compute $V = kq/r$ for each charge, keep the signs, and add. No components, no angles.

That difference makes potential far easier to compute, and it produces results that look paradoxical at first. At the midpoint between equal and opposite charges, the potentials cancel to zero while the two fields point the same way and reinforce (see worked example 2). Zero potential never guarantees zero field, and zero field never guarantees zero potential. When you need a fast pairwise force check, the Coulomb's law calculator handles the arithmetic.

Where this sits in the AP courses

In AP Physics 2, this material spans Topic 10.3 (Electric Fields), Topic 10.4 (Electric Potential Energy), and Topic 10.5 (Electric Potential) in Unit 10, Electric Force, Field, and Potential, weighted at 15 to 18 percent. In AP Physics C: Electricity and Magnetism, fields live in Unit 8 (Electric Charges, Fields, and Gauss's Law) and potential gets its own Unit 9. The calculus-based course builds Gauss's law and continuous charge distributions on top of the point-charge results here.

On either exam, a four-function, scientific, or graphing calculator is allowed on both sections, and the AP Physics 2 equation sheet supplies $k = 9.0 \times 10^9\ \mathrm{N \cdot m^2/C^2}$ . What actually gets tested is the field versus potential distinction: ranking tasks, points where one quantity is zero and the other is not, and sign reasoning with $\Delta U_E = q\Delta V$ . The full reference sheet is at AP Physics 2 formulas.

Field and potential 0.30 m from a point charge

A point charge $q = +4.0 \times 10^{-9}\ \mathrm{C}$ sits at the origin. Find the electric field and the electric potential at a point 0.30 m away.

List knowns: $q = +4.0 \times 10^{-9}\ \mathrm{C}$ , $r = 0.30\ \mathrm{m}$ , $k = 9.0 \times 10^9\ \mathrm{N \cdot m^2/C^2}$ .
Compute the product $kq$ once, since both formulas need it: $kq = (9.0 \times 10^9)(4.0 \times 10^{-9}) = 36\ \mathrm{N \cdot m^2/C}$ .
Field: $E = kq/r^2 = 36/(0.30\ \mathrm{m})^2 = 36/0.090 = 400\ \mathrm{N/C}$ . The charge is positive, so the field points radially away from it.
Potential: $V = kq/r = 36/0.30 = 120\ \mathrm{V}$ . Positive charge, positive potential, no direction to assign.

$E = 400\ \mathrm{N/C}$ directed away from the charge; $V = +120\ \mathrm{V}$ . Note the field used $r^2 = 0.090\ \mathrm{m^2}$ while the potential used $r = 0.30\ \mathrm{m}$ .

Midpoint of two opposite charges: zero V, nonzero E

Charge $q_1 = +3.0 \times 10^{-9}\ \mathrm{C}$ is at $x = 0$ and charge $q_2 = -3.0 \times 10^{-9}\ \mathrm{C}$ is at $x = 0.20\ \mathrm{m}$ . Find the electric field and the electric potential at the midpoint, $x = 0.10\ \mathrm{m}$ .

The midpoint is $r = 0.10\ \mathrm{m}$ from each charge.
Field from $q_1$ : $E_1 = kq/r^2 = (9.0 \times 10^9)(3.0 \times 10^{-9})/(0.10)^2 = 27/0.010 = 2700\ \mathrm{N/C}$ , pointing away from the positive charge, in the $+x$ direction.
Field from $q_2$ : same magnitude, $E_2 = 2700\ \mathrm{N/C}$ , pointing toward the negative charge, which is also the $+x$ direction. The two fields reinforce.
Add as vectors along $x$ : $E_{net} = 2700 + 2700 = 5400\ \mathrm{N/C}$ in the $+x$ direction.
Potential adds as signed numbers: $V = kq_1/r + kq_2/r = 270\ \mathrm{V} + (-270\ \mathrm{V}) = 0\ \mathrm{V}$ .

$E_{net} = 5400\ \mathrm{N/C}$ toward the negative charge, while $V = 0$ . Zero potential with a large field at the same point, exactly the trap AP questions set.

Speed gained from a potential difference

A small ball with charge $q = +2.0 \times 10^{-6}\ \mathrm{C}$ and mass $m = 0.0050\ \mathrm{kg}$ is released from rest at point A, where $V_A = 500\ \mathrm{V}$ . It moves to point B, where $V_B = 100\ \mathrm{V}$ . Ignoring gravity, how fast is it moving at B?

Potential difference: $\Delta V = V_B - V_A = 100 - 500 = -400\ \mathrm{V}$ .
Change in potential energy: $\Delta U_E = q\Delta V = (2.0 \times 10^{-6}\ \mathrm{C})(-400\ \mathrm{V}) = -8.0 \times 10^{-4}\ \mathrm{J}$ .
Energy conservation with only the electric force acting: $\Delta K = -\Delta U_E = +8.0 \times 10^{-4}\ \mathrm{J}$ . Starting from rest, $K_B = 8.0 \times 10^{-4}\ \mathrm{J}$ .
Solve $K = \frac{1}{2}mv^2$ for speed: $v = \sqrt{2K/m} = \sqrt{2(8.0 \times 10^{-4})/0.0050} = \sqrt{0.32} = 0.566\ \mathrm{m/s}$ .

$v \approx 0.57\ \mathrm{m/s}$ . The positive charge fell toward lower potential, lost $8.0 \times 10^{-4}\ \mathrm{J}$ of potential energy, and gained that much kinetic energy.

Frequently asked questions

What is the difference between electric field and electric potential?

Electric field (E = kq/r^2) is a vector measured in N/C: the force each coulomb of charge would feel, with a direction. Electric potential (V = kq/r) is a scalar measured in volts: the potential energy each coulomb would have, with a sign but no direction. Field falls off with the square of distance; potential falls off with the first power.

Can the electric potential be zero where the electric field is not zero?

Yes. At the midpoint between equal and opposite charges, the two potentials (+kq/r and -kq/r) cancel to zero, but the two fields point the same way and add. Zero potential just means a charge placed there would have zero potential energy relative to infinity; it says nothing about the force it would feel.

Is electric potential the same as electric potential energy?

No. Potential V is a property of a location in space, measured in volts (joules per coulomb), and exists whether or not any charge sits there. Potential energy U_E = qV is a property of a specific charge placed at that location, measured in joules. Moving a charge through a potential difference changes its energy by q times delta V.

Do I plug the sign of the charge into E = kq/r^2?

The cleanest habit is no: use the magnitude of q to get the field strength, then assign the direction by inspection (away from positive, toward negative). For potential the opposite habit applies: keep the sign of q, because V = kq/r is a signed scalar and negative charges genuinely produce negative potential.

What are the units of electric field and electric potential?

Field is measured in newtons per coulomb (N/C), which is exactly equivalent to volts per meter (V/m). Potential is measured in volts, where 1 V = 1 J/C. The V/m equivalence is a useful check: field tells you how fast potential changes with distance.